A Physics question with no clue or idea to the answer.
A box is dropped from an aeroplane 2000m high travelling horizontally at 100m/s. The box takes 20.2s to hit the ground. While the box speeds up vertically, it continues at 100m/s horizontally.What is the angle between the horizontal and the displacement of the box from the location it was released? Give your answer to 3 significant figures.Any help would be appreciated and any solutions would be even more so appreciated.
Thank you, I’ve found the solution
The length H should be 2039.803912
Root(100^2 +100^2)= approximately 141
Root ((144*20.20)^2 - 2000^2) = approximately 2039
And so I used SohCahToa to find angle theta which is 44.43548687
The length H should be 2039.803912
Root(100^2 +100^2)= approximately 141
Root ((144*20.20)^2 - 2000^2) = approximately 2039
And so I used SohCahToa to find angle theta which is 44.43548687
Original post by HarrelVeyson
Thank you, I’ve found the solution
The length H should be 2039.803912
Root(100^2 +100^2)= approximately 141
Root ((144*20.20)^2 - 2000^2) = approximately 2039
And so I used SohCahToa to find angle theta which is 44.43548687
The length H should be 2039.803912
Root(100^2 +100^2)= approximately 141
Root ((144*20.20)^2 - 2000^2) = approximately 2039
And so I used SohCahToa to find angle theta which is 44.43548687
What about horizontal distance = horizontal speed x time?
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